It is known that the solution set of inequality KX ^ 2 + 2kx-1 > 0 about X is an empty set, and the value range of real number k is obtained

It is known that the solution set of inequality KX ^ 2 + 2kx-1 > 0 about X is an empty set, and the value range of real number k is obtained

K=0
Then - 1 > 0
Satisfy that the solution set is an empty set
k≠0
If the solution set is empty, then KX & # 178; + 2kx-1
When k = 0, & nbsp; - 1 & gt; 0 is an empty set, when k ≠ 0, K & lt; 0, (2k) ^ 2 + 4K & lt; 0, the solution is & nbsp; - 4 & nbsp; & lt; & nbsp; K & lt; 0, which is an empty set, so the value range of K is (- 4,0] & nbsp;
Satisfy the constraint condition (absolute value x) + 2 (absolute value y)
|x|+2|y|
2cos 30 degree - 2 root sign 2sin60 degree multiplied by cos45 degree
The process must be complete
2cos30-2√2sin60×cos45
=2(cos30-√2sin60×cos45)
=2(√3/2-√2×√3/2×√2/2)
=2(√3/2-√3/2)
=0
cos30=√3/2
sin60=√3/2
cos45=√2/2
The original formula = √ 3-2 √ 2 * √ 3 / 2 * √ 2 / 2 = √ 3 - √ 3 = 0
Let a be a square matrix of order n. if AX = 0 is satisfied for any n-dimensional vector x, then a = 0?
incorrect
Yes | a ≠ 0
We know that AX = 0 has only zero solution, which is equivalent to | a | ≠ 0
As shown in the figure: AB and AC are chords, OM ⊥ AB is in M, on ⊥ AC is in N, is mn the median line of △ ABC?
yes,
Because AB and AC are chords, OM ⊥ AB is in M, and on ⊥ AC is in N, so m and N are the midpoint of AB and AC respectively, so Mn is the median line of △ ABC
What is the sum of binomial coefficients in the expansion of (2x-3y) to the 9th power?
Because there are many symbols, you can't type them here. The answer is as shown in the figure
Calculation of 2sin60 °− 2cos45 °− 3tan30 ° + tan45 °
The original formula is 2 × 32-2 × 22-3 × 33 + 1 = 3-1-3 + 1 = 0
It is known that a is a real symmetric matrix of order n. for any n-dimensional vector x, X '(transpose of x) AX = 0. It is proved that a = 0
The title is xtax = 0
What I said upstairs is wrong. A is a 0 matrix. How can I multiply by the inverse of a? Isn't that nonsense?
First of all, if a is a real symmetric matrix of order n, then a must be similar to a diagonal matrix. Let B = P ^ (- 1) AP, P ^ (- 1) be the inverse of P, then a = PBP ^ (- 1). For any n-dimensional vector x, x'ax = 0, then we can deduce that all diagonal elements of B are 0, that is, B = 0. According to a = PBP ^ (- 1), we can know that a = 0
Let x = (1,0,...) The first column of a is 0, and x = (0,1,...) 0) substituting into the second column of a, we can get a = 0
Because for any x, if you take x as the unit vector, ax = 0, and right multiply the inverse of a, then a = 0
Well, give it to me,
As shown in the figure, De is the median line of △ ABC, M is the midpoint of De, and the extension line of CM intersects AB at n, then nm: MC=______ .
∵ De is the median line of △ ABC, M is the midpoint of De, ∵ DM ∥ BC, DM = me = 14bc. ∵ NDM ∥ NBC, dmbc = nmcn = 14. ∥ NMMC = 13
Binomial (1-2x) power 2N-1, binomial coefficient is the largest term
In the expansion of (1-2x) ^ (2n-1), the term with the largest binomial coefficient is
c(2n-1,n-1)(-2x)^(n-1),
And C (2n-1, n) (- 2x) ^ n